是否可以在sqldf中使用典型于SQL Server 2008 DATEPART()的SQL命令？
我正在扫描文档，但未找到任何相关文档，因此我不熟悉SQLite，因此，如果我采用这种方式阅读，我将
我想做一些简单的事情，例如：
sqldf("select DISCHARGE_UNIT,
round(avg(datepart(hour,order_time)),2) `avg order time`
from data
group by DISHCHARGE_UNIT)
编辑
summary(data)
DISCHARGE.UNIT
4SOU :1295
2CAD :1250
2NOR :1185
4NOR :1144
3NOR :1125
3SOU :1080
(Other):2723
ORDER.DATE ORDER.TIME
1800-01-01:1213 12:00:00 AM:1213
1/4/2013 : 39 11:42:00 AM: 34
12/5/2012 : 36 11:51:00 AM: 34
2/15/2013 : 35 11:03:00 AM: 32
12/19/2012: 33 10:32:00 AM: 29
10/25/2012: 31 11:15:00 AM: 29
(Other) :8415 (Other) :8431
dput(head(data))
ORDER.TIME = structure(c(734L, 118L,
279L, 176L, 268L, 188L), .Label = c("1:00:00 PM", "1:01:00 PM",
"1:02:00 PM", "1:03:00 PM", "1:04:00 PM", "1:05:00 PM", "1:06:00 PM",
"1:07:00 PM", "1:08:00 PM", "1:09:00 PM", "1:10:00 PM", "1:11:00 PM",
"1:12:00 PM", "1:13:00 PM", "1:14:00 PM", "1:15:00 PM", "1:16:00 PM",
"1:17:00 PM", "1:18:00 PM", "1:19:00 PM", "1:20:00 PM", "1:21:00 PM",
"1:22:00 PM", "1:23:00 PM", "1:24:00 PM", "1:25:00 PM", "1:26:00 PM",
"1:27:00 PM", "1:28:00 PM", "1:29:00 PM", "1:30:00 PM", "1:31:00 PM",
"1:32:00 PM", "1:33:00 PM", "1:34:00 PM", "1:35:00 PM", "1:36:00 PM",
"1:37:00 PM", "1:38:00
ORDER.DATE = structure(c(297L, 352L, 340L, 299L, 400L, 185L
), .Label = c("1/1/2013", "1/10/2013", "1/11/2013", "1/12/2013",
"1/13/2013", "1/14/2013", "1/15/2013", "1/16/2013", "1/17/2013",
"1/18/2013", "1/19/2013", "1/2/2013", "1/20/2013", "1/21/2013",
"1/22/2013", "1/23/2013", "1/24/2013", "1/25/2013", "1/26/2013",
"1/27/2013", "1/28/2013", "1/29/2013", "1/3/2013", "1/30/2013",
"1/31/2013", "1/4/2013", "1/5/2013", "1/6/2013", "1/7/2013",
"1/8/2013", "1/9/2013", "10/1/2012", "10/1/2013", "10/10/2012",
"1
谢谢，
最佳答案
DATEPART(...)是Microsoft和Sybase专有的Transact-SQL的一部分，因此，不，sqldf不支持它。sqldf通过基于drv=参数中定义的任何引擎创建临时数据库来工作。因此，该引擎支持的任何SQL版本都应由sqldf支持。请参阅文档以获取受支持引擎的列表。如果未指定引擎，则它支持SQLite中定义的词汇，该词汇是标准SQL语言（documentation here）的子集。
您的示例中的一个问题是您的ORDER_TIME值使用12小时时间格式。 SQLite支持ISO-8601，它需要24小时格式。因此，在您的特定情况下，我唯一想到的选择是在R中创建一个HOURS列，并在对sqldf(...)的调用中使用它。
library(sqldf)
# create sample data...
set.seed(1)
# ORDER_TIME in 12-hour format; does not conform with ISO-8601
data <- data.frame(ORDER_TIME=strftime(as.POSIXct("1:00:00",format="%H:%M:%S")+seq(12*3600,24*3600,60),format="%I:%M:%S %p"),
ORDER.DATE=strftime(as.Date("2013-01-01")+rep(0:30,each=721),format="%m/%d/%Y"),
DISCHARGE_UNIT=sample(1:10,721,replace=T))
# add hours column based on ORDER_TIME
data$HOURS <- as.numeric(strftime(as.POSIXct(data$ORDER_TIME,format="%I:%M:%S %p"),format="%H"))
sqldf("select DISCHARGE_UNIT,
round(avg(HOURS),2) `avg order time`
from data
group by DISCHARGE_UNIT")
# DISCHARGE_UNIT avg order time
# 1 1 16.91
# 2 2 16.69
# 3 3 16.64
# 4 4 17.10
# 5 5 15.78
# 6 6 15.01
# 7 7 17.09
# 8 8 15.12
# 9 9 17.68
# 10 10 17.17
如果您的时间数据是24小时格式（强烈建议您这样做），则可以使用SQLite date and time functions：
set.seed(1)
# ORDER_TIME in 24-hour format; conforms to ISO-8601
data <- data.frame(ORDER_TIME=strftime(as.POSIXct("1:00:00",format="%H:%M:%S")+seq(12*3600,24*3600,60),format="%H:%M:%S"),
ORDER.DATE=strftime(as.Date("2013-01-01")+rep(0:30,each=721),format="%m/%d/%Y"),
DISCHARGE_UNIT=sample(1:10,721,replace=T))
sqldf("select DISCHARGE_UNIT,
round(avg(strftime('%H',ORDER_TIME)),2) `avg order time`
from data
group by DISCHARGE_UNIT",drv="SQLite")
# DISCHARGE_UNIT avg order time
# 1 1 16.91
# 2 2 16.69
# 3 3 16.64
# 4 4 17.10
# 5 5 15.78
# 6 6 15.01
# 7 7 17.09
# 8 8 15.12
# 9 9 17.68
# 10 10 17.17
关于r - sqldf中的DATEPART()，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/21341089/